Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?
Soluion : V=50 cm F=20cm
From mirror formula 1/u = 1/f – 1/v
= 1/20+ 1/50 =3/100 U = 33.3 cm
Therefore the distance of the object from the wall x = 50 – u
X = 50 – 33.3 = 16.7 cm.
Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced?
Answer: Here f = 15 cm, u = 40 cm
Now 1/f = 1/u + 1/v
Then 1/v = 1/f – 1/u
Or V= uf/uf = 40 X 15/25 = 24 cm
Then object is displaced towards the mirror let u^{1} be the distance object from the
Mirror in its new position.
Then u1 = (4020) = 20cm
If the image is formed at a distance u^{1} from the mirror then
v^{1 }= u^{1}f/u^{1}f = 20X15/20+15 = 60 cm. =  20 x15/20+15 = 60 cm.
Therefore the image will move away from the concave mirror through a distance equal to 60 – 24 = 36 cm.
Here u = 25 cm , v = 5 cm from the mirror formula 1/f =1/u + 1/v
Then 1/f = 1/25 + 1/5 = 4/ 25
F = 6.25 cm
As the focal length is positive the mirror is convex in nature.
Q. An object is placed in front of a convex mirror of radius of curvature 40 cm at a distance of 10 cm. Find the position, nature and magnification of mirror. Answer:
Q. An object is kept in front of a concave mirror of focal length of 15 cm. the image formed is 3 times the size of the object. Calculate the two possible distances of the object from the mirror.
Answer:
Q.A converging and a diverging of equal focal lengths are placed coaxially in Contact. Find the focal length and power of the combination.
Power of the combination p = 1/F = 0.
Practice Questions
Then 1/f = 1/25 + 1/5 = 4/ 25
F = 6.25 cm
As the focal length is positive the mirror is convex in nature.
Q. An object is placed in front of a convex mirror of radius of curvature 40 cm at a distance of 10 cm. Find the position, nature and magnification of mirror. Answer:
Here u = 10 cm, R = 40 cm
Then f = R/2 =  20 cm
From the mirror formula 1/v = 1/f – 1/ u = 1/20 + 1/10 = 1/20.
V= 20 cm so v is positive , a virtual and erect image will be formed on the other side of the object, i.e; behind the mirror. M=v/u = 20/10 = 2
Then f = R/2 =  20 cm
From the mirror formula 1/v = 1/f – 1/ u = 1/20 + 1/10 = 1/20.
V= 20 cm so v is positive , a virtual and erect image will be formed on the other side of the object, i.e; behind the mirror. M=v/u = 20/10 = 2
Answer:
Case:1. Image is real. M = 3
Here f= 15 cm
Now m=v/u = 3
Or , V = 3u
From the mirror formula
1/f = 1/u+1/v
1/15 = 1/u + 1/3u
U=  20 cm.
Case:2. When the image is virtual m = 3
Now m = v/u = 3
Or, V=3
Here f= 15 cm
Now m=v/u = 3
Or , V = 3u
From the mirror formula
1/f = 1/u+1/v
1/15 = 1/u + 1/3u
U=  20 cm.
Case:2. When the image is virtual m = 3
Now m = v/u = 3
Or, V=3
From the mirror formula
1/f=1/u+1/v
Then 1/15 = 1/u1/3u
2/3u = 1/15
U=10cm.
1/f=1/u+1/v
Then 1/15 = 1/u1/3u
2/3u = 1/15
U=10cm.
Q. Refractive index of
glass is 1.5 and that of water is 1.3. if the speed of light in water is
2.25X10^{8}m/s. What is the speed of light in glass?
Answer:
Here
n_{g}=1.5 and n_{w}=1.3 let v_{1} and v_{2} be the speeds of light in
glass and water respectively. If c is the speed of light in air then
c/v_{1}=1.5
and c/v_{2}=1.3
then
v_{1}/v_{2}= 1.3/1.5
v_{1}=1.3/1.5X2.25X10^{8}
^{}
v_{1}=1.95X10^{8}
m/s
Q. A light of wave
length 6000A^{0} in air enters a medium with refractive index 1.5.
what will be the frequency and wave length of light in medium?
Answer:
Here
wave length of light in air l=6000A^{0}
= 6X10^{7}m
Refractive
index of medium = 1.5 = n
The
frequency of light does not change, when light travels from air to a refracting
medium.
n=c/l = 3X10^{8}/6X10^{7}
= 5X10^{14}Hz
The
wave length of light in the medium l^{1}=l/n
= 6000/1.5 = 4000A^{0}
Q. Convex lens made up
of glass of refractive index 1.5 is dipped in turn in
(i) Medium
A of n=1.65
ii) Medium B of n = 1.33.
Explain giving reasons Whether
it will behave as a converging lens or diverging lens in each of the
Answer:
Two
cases.
Here n_{g}
Let f_{air} be the focal
length of lens in air then
1/f_{air} = (n_{g}1)(1/R_{1}1/R_{2})
(1/R_{1}1/R_{2}) = 1/f_{air}(n_{g}1)
= 2/f_{air……………………………..(1)}
(i) When
lens is dipped in a medium A here n_{A} = 1.65
Focal
length be f_{A} when dipped in a medium A then 1/f_{A}=(n_{A}1)(1/R_{1}1/R_{2})
Using
equation (1) we have
1/f_{A}
= (1.5/1.6 51) X 2/f_{air} = 1/5.5 f_{air}
F_{A}
= 5.5 f_{air}
As
sign of f_{A} is opposite to that of f_{air}, the lens will
behave as diverging lense.
(ii) When
lens is dipped in a Medium B n_{B} = 1.33
Let
f_{B} be the focal length of lens when dipped in medium B
Then
1/f_{B} = (n_{g}1)(1/R_{1}1/R_{2}) =(n_{g}/n_{B}1)
)(1/R_{1}1/R_{2})
1/f_{B
}=(1.5/1.331)X2/f_{air} = 0.34/1.33 f_{air} = F_{B }= 3.91
f_{air} as the Sign of f_{B} is same as that of f_{air} the lens will
behave as a converging lens
Q. A convergent beam
of light passes through a diverging lens of focal length 0.2 meters comes
to focus at a distance 0.3 meters behind the lens find the position of the
point at which the beam would converge in the absence of lens?
Answer: F=0.2
m, v=0.3 m
From
the lens equation 1/f = 1/v1/u
1/u
= 1/v1/f = 1/0.31/0.2 = 50/6
U=6/50
= 0.12 m
In
the absence of lens the beam would converge at a distance 0.12 m from the
present position of the lens.
Q. A beam of light
converges to a point P. A lens is placed in the path of convergent beam 12
cm from the point P. At what point the beam converges if the lens is (a) a
concave lens of focal length 16 cm (b) a convex length of focal length 20
cm.
Answer: (a)
here u =12cm f = 16 cm the lens equation we have
V=uf/u+f=12X16/1216
= 48 cm
as v is positive the beam converges on the same side that of point P
b)
Here u = 12 cm , f =20 cm from the lens equation we have
V=uf/u+f
= 12X20/12+20 = 240/32 = 7.5 cm
As v is positive the beam converges on the same
side as that of point of P
Answer: Let f and –f are be the focal length
of the converging and diverging lens respectively then focal length of the
combination
1/F = 1/f – 1/f = 0
Practice Questions
1. An object is placed in front of a concave mirror of radius of curvature 15cm at a distance of (a) 10cm. and (b) 5cm. Find the position, nature and magnification of the image in each case.
2. An object is placed 15cm from a concave mirror of radius of curvature 60 cm. Find the position of image and its magnification?
3. An object is kept at a distance of 5cm in front of a convex mirror of focal length 10cm. Give the position, magnification and the nature of the image formed.
4. An object is placed at a distance of 50cmfrom a concave lens of focal length 20cm. Find the nature and position of the image.
5. The power of a lens is 2.5 dioptre. What is the focal length and the type of lens?
6. What is the power of a concave lens of focal length 50cm?
7. Find the power of a concave lens of focal length 2m.
8. Two lens of power +3.5D and 2.5D are placed in contact. find the power and focal length of the lens combination.
9. A convex lens has a focal length of 20 cm. Calculate at what distance from the lens should an object be placed so that it forms an image at a distance of 40cm on the other side of the lens. State the nature of the image formed?
10. A 10cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30cm. The distance of the object from the line is 20cm.find the i)position ii)nature and iii) size of the image formed.
11. Find the focal length of a line power is given as +2.0D.
12. With respect to air the refractive index of ice and rock salt benzene are 1.31 and 1.54 respectively. Calculate the refractive index of rock salt with respect to ice.
13. An object 5cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.
14. The far point of a myopic person is 150cm in front of the eye. Calculate the focal length and the power of the lens required to enable him to see distant objects clearly.
2. An object is placed 15cm from a concave mirror of radius of curvature 60 cm. Find the position of image and its magnification?
3. An object is kept at a distance of 5cm in front of a convex mirror of focal length 10cm. Give the position, magnification and the nature of the image formed.
4. An object is placed at a distance of 50cmfrom a concave lens of focal length 20cm. Find the nature and position of the image.
5. The power of a lens is 2.5 dioptre. What is the focal length and the type of lens?
6. What is the power of a concave lens of focal length 50cm?
7. Find the power of a concave lens of focal length 2m.
8. Two lens of power +3.5D and 2.5D are placed in contact. find the power and focal length of the lens combination.
9. A convex lens has a focal length of 20 cm. Calculate at what distance from the lens should an object be placed so that it forms an image at a distance of 40cm on the other side of the lens. State the nature of the image formed?
10. A 10cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30cm. The distance of the object from the line is 20cm.find the i)position ii)nature and iii) size of the image formed.
11. Find the focal length of a line power is given as +2.0D.
12. With respect to air the refractive index of ice and rock salt benzene are 1.31 and 1.54 respectively. Calculate the refractive index of rock salt with respect to ice.
13. An object 5cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.
14. The far point of a myopic person is 150cm in front of the eye. Calculate the focal length and the power of the lens required to enable him to see distant objects clearly.
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Class X Science TermII

2 comments:
The far point of a myopic person is 150 cm in front of the eye. Calculate the focal length and the power of a lens required to enable him to see distant objects clearly.
Solution:
u = infinity
v = 150cm
1/f = 1/v  1/u
f = 150cm = 1.5m
P = 1/1.5 = 10/15 = 2/3 = 0.66 D
An object 5 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Soln.: Given h = 5 cm, u =20 cm, R = +30 cm or f = R/2 = 30 /2 = +15 cm, v = ? h = ?
Using 1/f = 1/v  1/u
we get v = 8.6 cm
The image is formed 8.6 cm behind the mirror. Thus the image is virtual and erect.
Now m = v/u =hi/ho
hi = v/u x ho = 8.6/20 x 5 = 2.15 cm
Thus the size of the image is 2.15 cm or 2.2. cm. The image is reduced.
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